Solve the following equation$x^{2}r^{2}+2r(2q-p) x+(p-2q)^{2}=0$

Given:  $x^{2}r^{2}+2r(2q-p) x+(p-2q)^{2}=0$

To do: Solve the given equation:

Solution:

 $x^{2}r^{2}+2r(2q-p) x+(p-2q)^{2}=0$

$(xr)^{2} - 2 \times (xr) \times (p - 2q) + (p - 2q)^{2}= 0$

Using identity $a^{2}- 2ab + b^{2} = (a - b)^{2}$ 

$(xr - (p - 2q))^{2} = 0$

$(xr - p + 2q)(xr - p + 2q)=0$

or $xr = p - 2q$ or $x = \frac{p - 2q}{r}$  Answer